Linear Algebra Examples

$y = \ln (x^{2} + \sqrt{x}) + a r c t g (\frac{e^{x}}{x}) + \frac{1}{x}$

Step 1

Set the argument in $\ln (x^{2} + \sqrt{x})$ greater than $0$ to find where the expression is defined.

$x^{2} + \sqrt{x} > 0$

Step 2

Solve for $x$ .

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Step 2.1

Subtract $x^{2}$ from both sides of the inequality.

$\sqrt{x} > - x^{2}$

Step 2.2

To remove the radical on the left side of the inequality, square both sides of the inequality.

${\sqrt{x}}^{2} > {(- x^{2})}^{2}$

Step 2.3

Simplify each side of the inequality.

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Step 2.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

${(x^{\frac{1}{2}})}^{2} > {(- x^{2})}^{2}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Simplify ${(x^{\frac{1}{2}})}^{2}$ .

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Step 2.3.2.1.1

Multiply the exponents in ${(x^{\frac{1}{2}})}^{2}$ .

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Step 2.3.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{\frac{1}{2} \cdot 2} > {(- x^{2})}^{2}$

Step 2.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.3.2.1.1.2.1

Cancel the common factor.

$x^{\frac{1}{2} \cdot 2} > {(- x^{2})}^{2}$

Step 2.3.2.1.1.2.2

Rewrite the expression.

$x^{1} > {(- x^{2})}^{2}$

Step 2.3.2.1.2

Simplify.

$x > {(- x^{2})}^{2}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Simplify ${(- x^{2})}^{2}$ .

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Step 2.3.3.1.1

Apply the product rule to $- x^{2}$ .

$x > {(- 1)}^{2} {(x^{2})}^{2}$

Step 2.3.3.1.2

Raise $- 1$ to the power of $2$ .

$x > 1 {(x^{2})}^{2}$

Step 2.3.3.1.3

Multiply ${(x^{2})}^{2}$ by $1$ .

$x > {(x^{2})}^{2}$

Step 2.3.3.1.4

Multiply the exponents in ${(x^{2})}^{2}$ .

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Step 2.3.3.1.4.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x > x^{2 \cdot 2}$

Step 2.3.3.1.4.2

Multiply $2$ by $2$ .

$x > x^{4}$

Step 2.4

Solve for $x$ .

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Step 2.4.1

Subtract $x^{4}$ from both sides of the inequality.

$x - x^{4} > 0$

Step 2.4.2

Convert the inequality to an equation.

$x - x^{4} = 0$

Step 2.4.3

Factor the left side of the equation.

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Step 2.4.3.1

Factor $x$ out of $x - x^{4}$ .

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Step 2.4.3.1.1

Raise $x$ to the power of $1$ .

$x - x^{4} = 0$

Step 2.4.3.1.2

Factor $x$ out of $x^{1}$ .

$x \cdot 1 - x^{4} = 0$

Step 2.4.3.1.3

Factor $x$ out of $- x^{4}$ .

$x \cdot 1 + x (- x^{3}) = 0$

Step 2.4.3.1.4

Factor $x$ out of $x \cdot 1 + x (- x^{3})$ .

$x (1 - x^{3}) = 0$

Step 2.4.3.2

Rewrite $1$ as $1^{3}$ .

$x (1^{3} - x^{3}) = 0$

Step 2.4.3.3

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = 1$ and $b = x$ .

$x ((1 - x) (1^{2} + 1 x + x^{2})) = 0$

Step 2.4.3.4

Factor.

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Step 2.4.3.4.1

Simplify.

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Step 2.4.3.4.1.1

One to any power is one.

$x ((1 - x) (1 + 1 x + x^{2})) = 0$

Step 2.4.3.4.1.2

Multiply $x$ by $1$ .

$x ((1 - x) (1 + x + x^{2})) = 0$

Step 2.4.3.4.2

Remove unnecessary parentheses.

$x (1 - x) (1 + x + x^{2}) = 0$

Step 2.4.4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$1 - x = 0$

$1 + x + x^{2} = 0$

Step 2.4.5

Set $x$ equal to $0$ .

$x = 0$

Step 2.4.6

Set $1 - x$ equal to $0$ and solve for $x$ .

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Step 2.4.6.1

Set $1 - x$ equal to $0$ .

$1 - x = 0$

Step 2.4.6.2

Solve $1 - x = 0$ for $x$ .

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Step 2.4.6.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 2.4.6.2.2

Divide each term in $- x = - 1$ by $- 1$ and simplify.

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Step 2.4.6.2.2.1

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Step 2.4.6.2.2.2

Simplify the left side.

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Step 2.4.6.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Step 2.4.6.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Step 2.4.6.2.2.3

Simplify the right side.

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Step 2.4.6.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 2.4.7

Set $1 + x + x^{2}$ equal to $0$ and solve for $x$ .

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Step 2.4.7.1

Set $1 + x + x^{2}$ equal to $0$ .

$1 + x + x^{2} = 0$

Step 2.4.7.2

Solve $1 + x + x^{2} = 0$ for $x$ .

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Step 2.4.7.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.4.7.2.2

Substitute the values $a = 1$ , $b = 1$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Step 2.4.7.2.3

Simplify.

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Step 2.4.7.2.3.1

Simplify the numerator.

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Step 2.4.7.2.3.1.1

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 2.4.7.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 2.4.7.2.3.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 2.4.7.2.3.1.2.2

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 2.4.7.2.3.1.3

Subtract $4$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 2.4.7.2.3.1.4

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 2.4.7.2.3.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 2.4.7.2.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 2.4.7.2.3.2

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2}$

Step 2.4.7.2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 2.4.7.2.4.1

Simplify the numerator.

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Step 2.4.7.2.4.1.1

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 2.4.7.2.4.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 2.4.7.2.4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 2.4.7.2.4.1.2.2

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 2.4.7.2.4.1.3

Subtract $4$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 2.4.7.2.4.1.4

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 2.4.7.2.4.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 2.4.7.2.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 2.4.7.2.4.2

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2}$

Step 2.4.7.2.4.3

Change the $\pm$ to $+$ .

$x = \frac{- 1 + i \sqrt{3}}{2}$

Step 2.4.7.2.4.4

Rewrite $- 1$ as $- 1 (1)$ .

$x = \frac{- 1 \cdot 1 + i \sqrt{3}}{2}$

Step 2.4.7.2.4.5

Factor $- 1$ out of $i \sqrt{3}$ .

$x = \frac{- 1 \cdot 1 - (- i \sqrt{3})}{2}$

Step 2.4.7.2.4.6

Factor $- 1$ out of $- 1 (1) - (- i \sqrt{3})$ .

$x = \frac{- 1 (1 - i \sqrt{3})}{2}$

Step 2.4.7.2.4.7

Move the negative in front of the fraction.

$x = - \frac{1 - i \sqrt{3}}{2}$

Step 2.4.7.2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 2.4.7.2.5.1

Simplify the numerator.

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Step 2.4.7.2.5.1.1

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 2.4.7.2.5.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 2.4.7.2.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 2.4.7.2.5.1.2.2

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 2.4.7.2.5.1.3

Subtract $4$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 2.4.7.2.5.1.4

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 2.4.7.2.5.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 2.4.7.2.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 2.4.7.2.5.2

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2}$

Step 2.4.7.2.5.3

Change the $\pm$ to $-$ .

$x = \frac{- 1 - i \sqrt{3}}{2}$

Step 2.4.7.2.5.4

Rewrite $- 1$ as $- 1 (1)$ .

$x = \frac{- 1 \cdot 1 - i \sqrt{3}}{2}$

Step 2.4.7.2.5.5

Factor $- 1$ out of $- i \sqrt{3}$ .

$x = \frac{- 1 \cdot 1 - (i \sqrt{3})}{2}$

Step 2.4.7.2.5.6

Factor $- 1$ out of $- 1 (1) - (i \sqrt{3})$ .

$x = \frac{- 1 (1 + i \sqrt{3})}{2}$

Step 2.4.7.2.5.7

Move the negative in front of the fraction.

$x = - \frac{1 + i \sqrt{3}}{2}$

Step 2.4.7.2.6

The final answer is the combination of both solutions.

$x = - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Step 2.4.8

The final solution is all the values that make $x (1 - x) (1 + x + x^{2}) = 0$ true.

$x = 0, 1, - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Step 2.5

Find the domain of $x^{2} + \sqrt{x}$ .

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Step 2.5.1

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

Step 2.5.2

The domain is all values of $x$ that make the expression defined.

$[0, \infty)$

Step 2.6

Use each root to create test intervals.

$x < 0$

$0 < x < 1$

$x > 1$

Step 2.7

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 2.7.1

Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Step 2.7.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 2.7.1.2

Replace $x$ with $- 2$ in the original inequality.

${(- 2)}^{2} + \sqrt{- 2} > 0$

Step 2.7.1.3

The left side is not equal to the right side, which means that the given statement is false.

False

Step 2.7.2

Test a value on the interval $0 < x < 1$ to see if it makes the inequality true.

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Step 2.7.2.1

Choose a value on the interval $0 < x < 1$ and see if this value makes the original inequality true.

$x = 0.5$

Step 2.7.2.2

Replace $x$ with $0.5$ in the original inequality.

${(0.5)}^{2} + \sqrt{0.5} > 0$

Step 2.7.2.3

The left side $0.95710678$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.7.3

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 2.7.3.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 2.7.3.2

Replace $x$ with $4$ in the original inequality.

${(4)}^{2} + \sqrt{4} > 0$

Step 2.7.3.3

The left side $18$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.7.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < 0$ False

$0 < x < 1$ True

$x > 1$ True

$x < 0$ False

$0 < x < 1$ True

$x > 1$ True

Step 2.8

The solution consists of all of the true intervals.

$0 < x < 1$ or $x > 1$

Step 3

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

Step 4

Set the denominator in $\frac{e^{x}}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 5

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(0, 1) \cup (1, \infty)$

Set-Builder Notation:

${x | x > 0, x \neq 1}$

Step 6